asymptotic curve on a complete surface.

Hello,

We have a 2-dimensional Riemannian Manifold immersed in , which is complete and has constant negative curvature.
Then the statement is, that for every point , there is an asymptotic curve , parametrized by arclength, s.t. c(0)=p.

I don't know how i can prove this. I mean the Surface M is complete, that is we can extend every geodesic into a geodesic on the whole real line .
And we know, locally there exists always a geodesic. But a geodesic is not an asymptotic curve. Since for an asymptotic curve we have that . But for a geodesic we have that c''(t) is perpendicular to .

So we have to search for another idea:
I could show, that for any point p, we have locally an asymptotic curve, parametrized by arclength.

But i have no idea, why we can extend such an asymptotic curve to the real line.....
I hope you can help me with this problem.

Re: asymptotic curve on a complete surface.

Yes you are right. I want to prove this theorem. And the author first assumes, that there is such an immersion of M, in the proof. (His strategy is proof by contradiction)
Then he says, that there must be such an asymptotic curve to each point.

Maybe we can do it like this:
We take an local asymptotic curve with . (as restriction of a regular curve on

Then we take the point and do the same process again with some function .
Now we define: by if x>=b and otherwise.

Re: asymptotic curve on a complete surface.

Since the Gaussian curvature is negative, the two principal curvatures k1 and k2, are of different signs. Let's say k1 > 0 and k2 < 0, then according to Euler's theorem, , this is a continuous function of with and , so there must be a value of so that k=0. This is the asymptotic direction.

So this defines a unit vector field X on the surface. At each point p, X(p) is the unit vector at the asymptotic direction. Since , and the principal directions are smooth on the surface, the above construction of X is smooth on the surface. Then at each p, there is locally a unique integral curve of X. Since M is complete, we can extend this integral curve indefinitely.